rary that this theory, dealt with and studied with profound attention, gives, at least in most cases, resistance absolutely zero; a singular paradox which I leave to geometricians to explain.(extract from Aerodynamics of T. Von Kerman)In order to get through the development of his theory and to explain where the error could have been. We first have to set the hypothesis with which we will work throughout the following derivation:-the flow is incompressible-the flow is non-viscous and then irrotational.-We will consider the flow around a cylindrical body.We consider a cylinder of radius R, with an imposed flow of velocity Ux far from the cylinder (see fig 1) fig 1We know that the fluid cannot penetrate the surface of the cylinder, so that the normal component of the velocity at the surface of the cylinder must be zero.In radial (polar) coordinates this is expressed as:nr(R,q) = 01.1We are now looking for the equation of n. Since we have seen that the flow was irrotational, we have: n = 01.2with called the del operator and expressed as: = /x I + /y j + /z k1.3and from vector calculus, we know that n = 0 (known as curl n) implies that n may be written as the gradient of a scalar function j : n = - j 1.4If we consider that the fluid is also incompressible (as it is the case here), then:.n = 0 (what is called the DIV of n)1.5from 1.4 and 1.5 we can have:.(-j) = j = 0.1.6It is important to note that j is a potential function solution of the Laplaces equation.The expression of the Laplaces equation is:j/x + j/y + j/z = 01.7and under its polar or radial form, for j(r,q):j/r + 1/r j/q + 1/r j/r = 01.8it has also been expressed as:1/r /r(r.j/r) + 1/r j/q = 01.9which means that j/r = 0 but why ?so in terms of j the boundary condition at infinity is j - Ux; in radial coordinates, x = r.cosq, so this becomes:j -U.r.cosq as r 1.10At surface of the cylinder we have:( j(r,q) / r )r=R = 01.11The strategy is that we will try to solve Laplac...
